next up previous contents
Next: 120 degree primitive triples,(a,b,c), Up: Pythagorean Triples, etc. Previous: Finding m and n   Contents


120 degree and 60 degree triples and the divisors 3, 5, and 7.

This section in pdf form. triples357.pdf



\includegraphics[width=2.5in]{../../../texdocs/ip4}


It's well known that if $ (a,b,c)$ is a a Pythagorean triple, that is, if $ (a,b,c)$ is a solution in positive integers to the 90 degree triangle equation $ a^2+b^2=c^2$ , then 3 and 4 each divides $ a$ or $ b$ , and 5 divides $ a, b$ or $ c$ where, of course, $ (3,4,5)$ is the smallest such solution.

A 120 degree triple, $ (a,b,c)$ , is a solution in positive integers to the 120 degree triangle equation

$\displaystyle a^2+b^2-2ab\cos120 ^{\circ}=a^2+b^2+ab=c^2.$

So, naturally, one wonders if a similar relationship exists between the positive integer solutions of 120 degree triangles and the smallest such solution, $ (3,5,7)$ ? It does, almost. To find such a relationship it's necessary to include an $ a+b$ term.


120 degree triples and the divisors 3, 5, and 7.


All primitive solutions to a 120 degree triple $ (a,b,c)$ , are given by the parametric equations:

$\displaystyle a=m^2-n^2,\quad b=2mn+n^2,$   and$\displaystyle \quad
c=m^2+n^2+mn.
$ (37)

where $ m$ and $ n$ are relatively prime, positive integers, $ m > n$ , and $ 3\nmid m-n$ . See (8) for a proof.


If $ (a,b,c)$ and $ (b,a,c)$ are considered the same solution, then the first 6 primitive solutions in order of smallest value for c are,

(1)
     $ 5^2+3^2+5\cdot 3=7^2$
(2)
     $ 8^2+7^2+8\cdot 7=13^2$
(3)
     $ 16^2+5^2+16\cdot 5=19^2$
(4)
     $ 24^2+11^2+24\cdot 11=31^2$
(5)
     $ 33^2+7^2+33\cdot 7=37^2$
(6)
     $ 35^2+13^2+35\cdot 13=43^2$

Notice that, in each case, 3 and 5, each, divides one of $ a, b$ , or $ a+b$ , and 7 divides one of $ a, b$ , $ a+b$ , or $ c$ .

Claim 8   If $ (a,b,c)$ is any 120 degree triple then 3 and 5 divides $ ab(a+b)$ , and 7 divides $ ab(a+b)c$ .

Proof. It's sufficient to show it's true for primitive triples. This claim can be proven directly by looking at residues modulo 3, 5, and 7; however it gets quite messy for the divisor 7. So, instead, I will use the parametric equations from (37) and the following result from Fermat's little theorem. That is, if $ s$ and $ t$ are integers, and $ p$ is a prime, then

$\displaystyle p  $   divides$\displaystyle    st\left(s^{p-1}-t^{p-1}\right).
$

From (37),

$\displaystyle ab(a+b)$ $\displaystyle =\left(m^2-n^2\right)\left(2mn+n^2\right)\left(m^2+2mn\right)$    
  $\displaystyle =m n\left(m^2 - n^2\right)\left(2\left(m^2 + n^2\right) - 5m n\right)$    
  $\displaystyle =2mn\left(m^4-n^4\right)-5\left(m^4n^2-m^2n^4\right) ,$    

and


$\displaystyle ab(a+b)c$ $\displaystyle =\left(m^2-n^2\right)\left(2mn+n^2\right)\left(m^2+2mn\right) \left(m^2+n^2+mn\right)$    
  $\displaystyle =2m n\left(m^6 - n^6\right) - 7\left(m^6 n^2 - m^5 n^3 + m^3 n^5 - m^2 n^6\right) .$    

Therefore, from Fermat's little theorem, 3 and 5 divide $ ab(a+b)$ , and 7 divides $ ab(a+b)c$ . And since 3, 5, and 7 are primes then 3 and 5 each divides one of $ a, b$ , or $ a+b$ , and 7 divides one of $ a, b,  a+b$ , or $ c$ . $ \qedsymbol$


60 degree triples and the divisors 3, 5, and 7.


A 60 degree triple, $ (p,q,r)$ , is a solution in positive integers to the 60 degree triangle equation

$\displaystyle p^2+q^2-2pq\cos60 ^{\circ}=p^2+q^2-pq=r^2.
$

Note that

$\displaystyle a^2+b^2+ab=(a+b)^2+b^2-(a+b)b=a^2+(a+b)^2-a(a+b).
$

Hence, if $ (a,b,c)$ is a 120 degree triple then $ (a+b,b,c)$ and $ (a,a+b,c)$ are 60 degree triples.


Here is a ``neat'' way to construct these three triangles.


\includegraphics[width=4.0in]{../../../texdocs/ip2}


On line $ l$ layout line segments $ AB$ and $ BE$ having lengths $ a$ and $ b$ respectively, where $ a$ and $ b$ are the side lengths of a 120 degree triangle. On and below $ AB$ construct equilateral triangle $ ADB$ with sides of length $ a$ . On and above $ BE$ construct equilateral triangle $ BEC$ with sides of length $ b$ . Hence $ \angle
ABD$ and $ \angle CBE$ are each 60 degrees. So point $ B$ lies on line segment $ DC$ and $ \angle ABC$ is 120 degrees. Draw line segment $ AC$ . Thus, the construction shows the 120 degree triangle $ ABC$ and its two associated 60 degree triangles $ AEC$ and $ ADC$ .


Let $ u^2+v^2-uv=w^2$ . If $ u, v$ , and $ w$ are positive integers, then $ (u,v,w)$ is a 60 degree triple. If, additionally, $ u, v$ , and $ w$ are pairwise relatively prime, then $ (u,v,w)$ is a primitive 60 degree triple. The first seven such triples in order of the smallest value for $ w$ are,

(1)
     $ 1^2+1^2-1\cdot1= 1^2$
(2)
     $ 8^2+5^2-8\cdot5= 7^2$
(3)
     $ 8^2+3^2-8\cdot 3=7^2$
(4)
     $ 15^2+7^2-15\cdot 7=13^2$
(5)
     $ 15^2+8^2-15\cdot 8=13^2$
(6)
     $ 21^2+5^2-21\cdot 5=19^2$
(7)
     $ 21^2+16^2-21\cdot 16=19^2$

Notice that, in each case, 3 and 5, each, divides one of $ u, v$ , or $ u-v$ , and 7 divides one of $ u, v$ , $ u-v$ , or $ w$ .

Claim 9   If $ (u,v,w)$ is any 60 degree triple then 3 and 5 divides $ uv(u-v)$ , and 7 divides $ uv(u-v)w$ .

Proof. It's sufficient to show the claim is true for primitive triples. Clearly it's true for the triple $ (1,1,1)$ . So let $ u^2+v^2-uv=w^2$ where $ (u,v,w)$ is a primitive triple, $ uvw\neq
1$ . Without loss of generality, let $ u$ be greater than $ v$ , then

$\displaystyle (u-v)^2+v^2+(u-v)v=u^2+v^2-uv=w^2.$

Hence $ (u-v,v,w)$ is a 120 degree triple. So, from claim (8),

$\displaystyle 3 \cdot 5 \mid (u-v)v\bigl((u-v)+v\bigr)=uv(u-v),$   and$\displaystyle \quad 7
\mid (u-v)v\bigl((u-v)+v\bigr)w=uv(u-v)w.
$

$ \qedsymbol$


The drawing below shows two 60 degree triangles $ AEC$ and $ ADC$ along with their associated 120 degree triangle $ ABC$ .


\includegraphics[width=4.3in]{../../../texdocs/ip3}


If the opposite side is to a power


Theorem 8   Let $ a$ and $ b$ be non-zero integers, and let $ c, k,$ and $ d$ be positive integers such that $ a^2+b^2+ab=\left(c^k\right)^2$ where $ d$ is a divisor of $ k$ and $ a, b,$ and $ c$ are pairwise relatively prime.

  1. If $ p=6d-1$ is a prime then $ p$ divides one of $ a, b,$ and $ a+b.$
  2. If $ q=6d+1$ is a prime then $ q$ divides one of $ a, b, a+b,$ and $ c.$
  3. If $ d=3^j$ , $ j$ a non-negative integer, then $ 3^{j+1}$ divides one of $ a, b,$ and $ a+b.$


To find relative prime non-zero integer solutions to $ a^2+b^2+ab=\left(c^k\right)^2$ , set

$\displaystyle a=\frac{\omega
(m-n\bar{\omega})^{2k}-\bar{\omega}(m-n\omega)^{2k...
...\mbox{and}\quad b=\frac{
(m-n\bar{\omega})^{2k}-(m-n\omega)^{2k}}{\sqrt{3} i}
$

where

$\displaystyle \omega=\frac{-1}{2}+\frac{1}{2} \sqrt{3} i,\quad \bar{\omega}=\frac{-1}{2}-\frac{1}{2} \sqrt{3} i,$   and$\displaystyle \quad i=\sqrt{-1} .
$

And where $ m$ and $ n$ are positive integers, $ m > n$ , and $ m-n$ is not a multiple of $ 3.$ Then $ c=m^2+n^2+mn$ .


Examples



Table 6: Divisors of 120 degree triples
Triangle $ k$ $ d$ Divisors of $ a, b,$ or $ a+b$ Divisors of $ a, b, a+b,$ or $ c$
$ a^2+b^2+ab=c^2$ 1 1 $ 3, 5$ $ 3, 5, 7$
$ a^2+b^2+ab=\left(c^2\right)^2$ 2 1,2 $ 3, 5, 11$ $ 3, 5, 7, 11, 13$
$ a^2+b^2+ab=\left(c^3\right)^2$ 3 1,3 $ 5, 3^2, 17$ $ 5, 7, 3^2, 17, 19$
$ a^2+b^2+ab=\left(c^4\right)^2$ 4 1,2,4 $ 3, 5, 11, 23$ $ 3, 5, 7, 11, 13, 23$
$ a^2+b^2+ab=\left(c^5\right)^2$ 5 1,5 $ 3, 5, 29$ $ 3, 5, 7, 29, 31$
$ a^2+b^2+ab=\left(c^6\right)^2$ 6 1,2,3,6 $ 5, 3^2, 11, 17$ $ 5, 7, 3^2, 11, 13, 17, 19, 37$
$ a^2+b^2+ab=\left(c^7\right)^2$ 7 1,7 $ 3, 5, 41$ $ 3, 5, 7, 41, 43$
$ a^2+b^2+ab=\left(c^8\right)^2$ 8 1,2,4,8 $ 3, 5, 11, 23, 47$ $ 3, 5, 7, 11, 13, 23, 47$
$ a^2+b^2+ab=\left(c^9\right)^2$ 9 1,3,9 $ 5, 17, 3^3, 53$ $ 5, 7, 17, 19, 3^3, 53$
$ a^2+b^2+ab=\left(c^{10}\right)^2$ 10 1,2,5,10 $ 3, 5, 11, 29, 59$ $ 3, 5, 7, 11, 13, 29, 31,59, 61$



Note that if integers $ a$ and $ b$ have the same sign then $ \vert a\vert+\vert b\vert$ and $ a+b$ have the same divisors. And if $ a$ and $ b$ have opposite parity then $ \vert a\vert+\vert b\vert$ and $ a-b$ have the same divisors. So to change table (6) to a table of divisors of 60 degree triples all that's necessary is to change $ a+b$ to $ a-b.$


next up previous contents
Next: 120 degree primitive triples,(a,b,c), Up: Pythagorean Triples, etc. Previous: Finding m and n   Contents
f. barnes 2008-04-29
Hosted by www.Geocities.ws

1